Question

In Exercises \(19 - 28 ,\) determine the limit graphically. Confirm algebraically. $$\lim _ { x \rightarrow 0 } \frac { \sin x } { 2 x ^ { 2 } - x }$$

Short answer

The limit as \(x\) approaches 0 for the function \(\frac { \sin x } { 2 x ^ { 2 } - x }\) does not exist both graphically and algebraically.

Step-by-step solution

Graphical Determination

Graph the given function \(\frac { \sin x } { 2 x ^ { 2 } - x }\) and observe the behavior of the function as the x-value approaches zero. It's worth noting that for \(x=0\), the function is undefined. However, we can look at the function around \(x=0\) instead. If the function approaches a particular value when x is very close to zero, that value is the limit.

Algebraic Determination

For algebraically determining the limit, we can't directly substitute \(x=0\) in the function because it would lead to division by zero. We can, however, rewrite the function to \(\frac { \sin x } { x(2x - 1) }\) and apply the limit theorem.

Apply Limit Theorem

Now apply the limit theorem to each part of the function. The limit of \(sinx\) as \(x\) approaches zero is zero. To determine the limit of the denominator as \(x\) approaches zero, directly substitute \(x=0\) to get 0. This gives us a form of \(0/0\) which is indeterminate. However, as \(x\) approaches zero from the right side, it's more rigorous to take the limit as \(x\) approaches zero from the right side, or \( \lim _ { x \rightarrow 0^{+}} \frac { x(2x - 1) }\). This limit happens to be 0. As the function changes behavior around \(x=0\), the limit as \(x\) approaches \(0\) does not exist as it is not the same from the left and right leading to an undefined behavior in the graph.